A377224 - OEIS

%I #20 Jan 08 2025 11:40:15

%S 1,0,1,1,2,1,3,1,2,3,2,3,2,2,1,3,1,3,4,1,3,2,4,2,6,2,4,5,4,3,5,3,3,4,

%T 2,2,4,1,3,3,3,3,7,1,6,6,6,3,8,4,3,7,3,7,4,4,2,4,1,5,6,1,6,7,4,4,9,6,

%U 5,8,3,6,5,3,4,5,3,3,4,1,9,6,5,3,9,5,6,9,6,8,10,3,3,9,4,7,7,4,7,5,4

%N Number of ways to write n as x*(5*x+1) + y*(5*y+1)/2 + z*(5*z+1)/2, where x,y,z are integers with y*(5*y+1) <= z*(5*z+1).

%C Conjecture 1: a(n) = 0 only for n = 1. Also, a(n) = 1 only for n = 0, 2, 3, 5, 7, 14, 16, 19, 37, 43, 58, 61, 79.

%C This has been verified for n <= 2*10^6.

%C Conjecture 2: Let N be the set of all nonnegative integers. Then

%C {x*(5*x+1) + y*(5*y+1)/2 + 5*z*(5*z+1)/2: x,y,z are integers} = N\{1,5},

%C {x*(5*x+1) + y*(5*y+1)/2 + 3*z*(5*z+1)/2: x,y,z are integers} = N\{1,5,32},

%C {x*(5*x+1) + y*(5*y+1)/2 + 2*z*(5*z+1): x,y,z are integers} = N\{1,5,70},

%C and

%C {x*(5*x+1)/2 + y*(5*y+1)/2 + z*(5*z+1)/2: x,y,z are integers} = N\{1,10,19,94}.

%C Conjecture 3: We have

%C {x*(5*x+3) + y*(5*y+3)/2 + 3*z*(5*z+3)/2: x,y,z are integers} = N\{31,77},

%C {x*(5*x+3) + y*(5*y+3)/2 + 5*z*(5*z+3): x,y,z are integers} = N\{10,16},

%C and

%C {x*(5*x+3)/2 + y*(5*y+3)/2 + 5*z*(5*z+3)/2: x,y,z are integers} = N\{3,15,29,44}.

%H Zhi-Wei Sun, <a href="/A377224/b377224.txt">Table of n, a(n) for n = 0..10000</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2015.10.014">A result similar to Lagrange's theorem</a>, J. Number Theory 162 (2016), 190-211.

%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-017-9354-4">Universal sums of three quadratic polynomials</a>, Sci. China Math. 63 (2020), 501-520.

%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/2411.14308">New results similar to Lagrange's four-square theorem</a>, arXiv:2411.14308 [math.NT], 2024.

%e a(14) = 1 with 14 = 0*(5*0+1) + 1*(5*1+1)/2 + 2*(5*2+1)/2.

%e a(37) = 1 with 37 = (-1)*(5*(-1)+1) + (-2)*(5*(-2)+1)/2 + 3*(5*3+1)/2.

%e a(58) = 1 with 58 = (-2)*(5*(-2)+1) + (-1)*(5*(-1)+1)/2 + (-4)*(5*(-4)+1)/2.

%e a(79) = 1 with 79 = -4*(5*(-4)+1) + 0*(5*0+1)/2 + 1*(5*1+1)/2.

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

%t tab={};Do[r=0;Do[If[SQ[40(n-x(5x+1)-y(5y+1)/2)+1],r=r+1],{x,-Floor[(Sqrt[20n+1]+1)/10],(Sqrt[20n+1]-1)/10},{y,-Floor[(Sqrt[20(n-x(5x+1))+1]+1)/10],Floor[(Sqrt[20(n-x(5x+1))+1]-1)/10]}];tab=Append[tab,r],{n,0,100}];Print[tab]

%Y Cf. A057569, A085787, A306383.

%K nonn

%O 0,5

%A _Zhi-Wei Sun_, Nov 13 2024